Master Organic Chemistry Reaction Guide

Addition of HBr to Alkenes

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Description: Treatment of alkenes with hydrobromic acid will result in the formation of alkyl bromides.

Notes: This is an addition reaction. Note that the bromine always ends up at the more substituted carbon of the alkene (Markovnikoff-selectivity)

Examples:

examples of addition of HBr to alkenes

Notes:  The first two examples show a simple addition of HBr to give the Markovnikov product. In the third example Markovnikov’s rule gives no clear preference, so a mixture is obtained.  The fifth example gives a mixture of diastereomers, since addition of HBr across the alkene will not affect the initial (R) stereochemistry.  The final (sixth) example will give a dominant product, despite identical substitution, because formation of the resonance-stabilized carbocation will be favored!

When a secondary carbocation is formed adjacent to a tertiary or quaternary carbon, rearrangements are possible. See these pages for more examples:

Mechanism: Electrons from the C1-C2 π bond attack the hydrogen of HBr, expelling the bromide anion and leading to the formation of a carbocation (Step 1, arrows A and B). Note here that the carbocation preferentially forms on C2 (secondary) and not C1 (primary) since secondary carbocations are more stable. The bromide anion then attacks the carbocation, leading to formation of the alkyl bromide (Step 2, arrow C)

Notes: Since the carbocation is planar (flat) there is no preferred direction of attack of the bromide ion. If it’s possible to form a stereocenter, a mixture of stereoisomers will be obtained.

Rearrangements: When secondary (or primary) carbocations are formed adjacent to a more substituted carbon, adjacent hydrogen atoms or alkyl groups can shift, resulting in formation of a more stable carbocation.

The first step of the reaction below is formation of a carbocation. Here, the formation of a carbocation via attack of the alkene upon H-Br is shown.  (Step 1, arrows A and B). Since we have a secondary carbocation adjacent to a tertiary carbon, shift of a hydrogen to the secondary carbocation will result in a (more stable) tertiary carbocation (Step 2, arrow C). The tertiary carbocation is then trapped, in this case, by Br(–) (Step 3, arrows D). The rearrangement step goes through a transition state such as the one pictured.

 

Additional example (advanced) : If the molecule contains adjacent stereocenters, the two directions of attack on the carbocation will no longer be of equal energy and a mixture of diastereomers will be obtained. In this example, attack of the bromide ion on the less hindered “top face” of the carbocation is favored, and a mixture of products (diastereomers) will be obtained.

Rearrangements can also occur. See Additions To Alkenes Accompanied By 1,2-Hydride Shifts

Quiz Yourself!

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(Advanced) References And Further Reading:

  1. Early example
    The Stereochemistry of the Addition of Hydrogen Bromide to 1,2-Dimethylcyclohexene
    George S. Hammond and Thomas D. Nevitt
    Journal of the American Chemical Society 1954 76 (16), 4121-4123
    DOI: 10.1021/ja01645a020
    Early paper from the 50’s by Prof. George Hammond (of Hammond’s Postulate) on the mechanism of HBr addition to 1,2-dimethylcyclohexane. He prefers a concerted pathway, although that might due to the conditions he employs – in pentane, a very nonpolar solvent, polar intermediates are disfavored.
  2. Mechanistic studies
    Hydrochlorination of cyclohexene in acetic acid. Kinetic and product studies
    Robert C. Fahey, Michael W. Monahan, and C. Allen McPherson
    Journal of the American Chemical Society 1970 92 (9), 2810-2815
    DOI: 10.1021/ja00712a034
    Detailed kinetic studies of the addition of HCl to cyclohexene in acetic acid, discussing a possible third-order mechanism (rate = k[cyclohexene][HX]2).
  3. Experimental Procedure

    SPIROANNELATION OF ENOL SILANES: 2-OXO-5-METHOXYSPlRO[5.4]DECANE

    Lee, T. V.; Porter, J. R. Org. Synth. 1995, 72, 189
    DOI: 10.15227/orgsyn.072.0189
    The first reaction in the above procedure involves two steps – addition of HBr across the double bond and converting the aldehyde to a dimethyl acetal.


Real-Life Examples:

Org. Synth. 1938, 18, 47

DOI Link: 10.15227/orgsyn.018.0047

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Org. Synth. 1940, 20, 64

DOI Link: 10.15227/orgsyn.020.0064

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Comments

Comment section

31 thoughts on “Addition of HBr to Alkenes

  1. This really is as good as anything out there- and I think the way it is neatly packaged makes it better than the rest. I am new as of yesterday to being a member- so if this is a frequently asked new kid question I apologize- but is there a clean way to print these? I filled out the mechanism blank sheets in Organic Chem as a Second Language and I wanted to print these out so that I can study from both my copy of notes and the reactions from this guide to make sure I have it all down- but printing it from a source like this takes a lot of pages since it prints out everything not just the text and pics about the mechanism (like this one short reaction printed is 9 pages). If not- that is okay- thanks for all of your hard work.

    1. Good question! The transition state involves partial bonds and non-ideal bonding geometries, which is less stable than the carbocation itself.

  2. This is not 100% related to this video, but on the topic of alkenes in general. Why is 3-methyl-2-butene more reactive than 2-methyl-1-butene?

  3. Will solvent change the product? Eg.- If water is used as a solvent will OH- attach instead of Br- as it is better nucleophile?

  4. Hi I was wondering its not mentioned on here but with the stereochemistry of the product formed. Which step in the mechanism determines the stereochemistry of the product? I can’t seem to figure it out :S Does it have something to do with which carbocation intermediate is formed?

    1. There are two steps in the mechanism which will determine the stereochemistry.

      In the first step, the alkene can attack H+ from either of its two faces. If this results in a chiral center, then this step will determine if this carbon is “R” or “S”.
      In the second step, the flat carbocation can be attacked from either face by the Br- nucleophile. If this results in a chiral center, then this step will determine if *this* carbon is “R” or “S”.
      So if two new chiral centers are formed, it’s possible to have 4 new products! (Thankfully this won’t always be the case).

    1. Yes, but you wouldn’t really see the result, since it will protonate the ring (forming C-H) and then break C-H to re-form the aromatic compound. So it looks as if no reaction is happening.
      However if you use D-Br what will happen is that there will be slow exchange of H for D at the aromatic ring.

  5. Let’s say you have a 3-methylcyclopentene reacted to HBr. How would it produce 1-bromo-1-methylcyclopentane? (It means the bromide ion became geminal to the methyl group)

  6. Hello sir, useful information.
    Can you please tell me, if there is option between a OH group and double bond, HBr will add to double bond or protonate OH group?

    1. Proton transfers to OH are generally faster than is addition to a double bond. However, it’s an equilibrium situation. If your molecule contains both an alcohol and an alkene, I’d double-check to make sure that you can’t have intramolecular formation of a 5 or 6 membered ring from 1) addition of HBr across the double bond, and 2) attack of the OH on the resulting carbocation.
      The other option is that the protonated OH leaves as OH2, resulting in a new carbocation, which is then attacked by a double bond as nucleophile. It really depends on your molecule.

  7. This is in regards to the “additional example (advanced)”: I am not sure that it is reasonable to use sterics on the a carbocation to predict which diastereomer is favored. Even if you are under kinetic control as your answer would require, depending on the side of the ring it attacks there are considerations beyond sterics, such as torsional strain developing at the transition state that can trump “same side of the ring” sterics.

    Lastly, bare alkenes and HX’s often follow a rate equation that is second order with respect to the HX. Therefore, the nucleophile is often delivered to the opposite face than the H+, as the developing carbocation is captured immediately as it is forming.

    Final thought: as an organic chemistry professor trying to get students to appreciate “selectivity vs reactivity” and that carbocations are highly reactive intermediates, I think it’s better (and more true) to note that there is a mixture of diastereomers formed that won’t be 50/50 (because they’re of different energy).

    1. For the radical pathway, the free-radical pathway for addition is only favorable for HBr. Radical addition of HCl is onlky rarely observed. Here is an example: https://pubs.acs.org/doi/abs/10.1021/ja00879a034 .
      The reason is that the HCl bond is too strong (103 kcal/mol) to be easily broken by the alkyl radical that forms after addition of HCl – a typical secondary C-H bond is about 96 kcal/mol, so the reaction is thermodynamically uphill.
      For HI, the problem is that the C-C pi bond (60 kcal/mol) is weaker than the C-I bond (50 kcal/mol) so there is no driving force for the addition reaction to occur.
      HF suffers from the same problem as HCl, but moreso – the HF bond is extremely strong. Even if you do make the F• radical, it’s nearly impossible to control.
      So to summarize HBr is kind of in the “goldilocks” region for this process. The C-Br bond is stronger (66 kcal/mol) than the C-C pi bond (60 kcal/mol), facilitating addition, and the C-H bond (96 kcal/mol for a secondary carbon) is stronger than the H-Br bond (87 kcal/mol) meaning C-H abstraction will be thermodynamically downhill.
      Bond strengths from here: https://labs.chem.ucsb.edu/zakarian/armen/11—bonddissociationenergy.pdf

  8. Sir,if hbr in ccl4 is used,will it change the addition reaction mechanism??What role will ccl4 actually play here??

    1. Just a non-polar organic solvent. It wouldn’t change the reaction mechanism. Although you’d want to make sure you don’t also see “light” or “peroxides” because that would indicate that you’re performing a free-radical addition of HBr across the alkene, which *does* involve a different mechanism.

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